Question

A group of business administration students conducted a survey on their college campus to find out student demand for a product, a protein supplement for fruit smoothies. One of the first steps was to extract a random sample of 280 students and obtain data that could be useful in developing their marketing strategy.

If {a ratio of 0.42 students decide to drink smoothies in the afternoon. Find the margin of error for the 90% confidence interval for the proportion of students who drink smoothies in the afternoon.

Select one:

to. the margin of error is E = 0.4200

b. The margin of error is E = 0.0017

c. the margin of error is E = 0.0578

d. The margin of error is E = 0.0014

and. The margin of error is E = 0.0485

Answer 1

Solution :

Given that,

Point estimate () = 0.42

1 - = 0.58

Z/2 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 * (((0.42 * 0.58) / 280)

= 0.0485

The margin of error is E = 0.0485